OPhO 2025, Финален кръг · част I — Задача 1. Theory Question 1

• (A.1) Boundary conditions give $\tan(\theta_1)/\tan(\theta_0) = \mu_r$, so $\theta_0 \to 0$ as $\mu_r \to \infty$.

• (A.2) $F_{\text{(ii)}} = \frac{\mu_0 I^2 \beta}{4\sqrt{2}\alpha}\left((\rho^2 - 1)J\left(\frac{\rho^4 - 2\beta^2 + 1}{2\alpha^2}\right) - (\rho^2 + 1)J\left(\frac{\rho^4 + 2\beta^2 + 1}{2\alpha^2}\right) - 2\rho^2 J\left(1 + \frac{2\beta^2}{\alpha^2}\right)\right),\quad \rho^2 = \alpha^2 + \beta^2$

  Answer of the "original solution"; the official Erratum states that the image-loop construction for (b)-(d) is in error and that full points for (b)-(d) were given for the key idea of the sphere-plane image setup.

• (A.3) 1.097

  At $\beta \approx 0.948$; from the original (erratum-affected) solution.

• (A.4) $\frac{1}{(\beta^2 - 1)^4} - \frac{1}{(\beta^2 + 1)^4} - \frac{1}{16\beta^3} = 0,\quad \beta \approx 2.079$

  From the original (erratum-affected) solution.

• (B.1) $\frac{dM}{dt} \approx \frac{16\pi\rho G^2 M^2}{c^2 v}$

• (B.2) The drag force diverges (logarithmically): $F = \frac{2\pi\rho G^2 M^2}{v^2}\int_1^\infty \frac{1}{u}\ du$; the assumption that particles arbitrarily far away are affected only by the black hole is not valid.

• (B.3) $F \approx \frac{4\pi\rho G^2 M^2}{3v^2}\ln\left(\frac{v^6}{\rho G^3 M^2}\right)$

Magnetosphere - Solution

(a) Consider the fields $\vec{B}_0, \vec{H}_0$ and $\vec{B}_1, \vec{H}_1$ outside and inside the iron respectively. Gauss's Law, $\nabla \cdot \vec{B} = 0$, shows that $B_{0,\perp} = B_{1,\perp}$ while Ampere's Law, $\nabla \times \vec{H} = \vec{J}_f = \vec{0}$, shows that $\vec{H}_{0,\parallel} = \vec{H}_{1,\parallel}$. These are the boundary conditions for magnetic materials. Using $\vec{H} = \vec{B}/\mu$, we have $\vec{B}_{1,\parallel} = \mu_r \vec{B}_{0,\parallel}$. Thus, if $\theta_0$ and $\theta_1$ are the angles to the normal, we have:

Assuming that $B_{1,\parallel}$ cannot become arbitrarily large, this implies that $\theta_0 \to 0$ as $\mu_r \to \infty$.

Marking Scheme: 0.25 for applying $\vec{B}$ and $\vec{H}$ boundary conditions. 0.25 for obtaining ”refraction” law relating $\theta_1$ and $\theta_2$.

Original solution for (b)-(d) (b) By uniqueness, it suffices to find a field $\vec{B}$ that is perpendicular to the interfaces and has the correct circulation around the wire, so we may use the method of images. Note that the image of the current loop in the plane is a congruent current loop with the same orientation.

To show that the image of the current loop in the sphere is another current loop, consider the equivalent configuration of a spherical cap with radius $\alpha R$ and magnetic dipole moment per unit area $I$ perpendicular to the cap. This lets us apply well-known results from electrostatics. The image of this cap is another cap with radius $\alpha' R = \frac{\alpha}{\rho^2} R$, where $\rho^2 = \alpha^2 + \beta^2$. Furthermore, the image of an area $dA$ has area $dA' = \frac{dA}{\rho^4}$ and dipole moment $dm' = \frac{dm}{\rho^3}$, giving $I' = \frac{dm'}{dA'} = I\rho$ (with the same orientation).

Thus, we can consider the following set of three image loops. Because the loops can be divided into image pairs either in the sphere or in the plane, the resulting field is perpendicular to both of these surfaces and thus satisfies the boundary conditions.

Next, we note that the force on the hemisphere is equal to the force on loop (ii) from the other three. This is because we could replace the hemisphere with loop (ii) without changing the external field, so the forces on loop (i) and on the plane would not change. Here, it is necessary to have an air gap between the hemisphere and the plane.

Thus, we need to find the force between coaxial current loops. Consider the general case of two loops with radii $R, aR$ and currents $I_1, I_2$ whose centers are at $z = 0, bR$. First, we'll find the magnetic vector potential of the first loop, working in cylindrical coordinates:

Here, the $\hat{r}$ component integrates to 0. Thus, the force on the second loop is:

Note that this is independent of $R$. Finally, we can write the total force on loop (ii) as:

(c) Plotting the region where $F_{\text{(ii)}} > 0$ in the $(\alpha, \beta)$ plane:

We see that the maximum value of $\alpha$ is $\boxed{\alpha \approx 1.097}$; here, $\beta \approx 0.948$.

(d) According to the plot, the maximum value of $\beta$ has $R = 0$, so we can treat the current rings as point dipoles. The field of a dipole $m$ along its axis is $B_z = \frac{\mu_0 m}{2\pi z^3}$, so the force on a dipole $m'$ located at $z = z_0$ is $F_z = m'\frac{dB_z}{dz} = -\frac{3\mu_0 m m'}{2\pi z_0^4}$. Letting ring (i) have moment $m$, ring (i$'$) has moment $m$ while rings (ii) and (ii$'$) have moments $\frac{m}{\beta^3}$. Thus, at the threshold we have:

This agrees with the plot.

Erratum Unfortunately, the image charges for part (b) are in error. The image of a dipole in a sphere is not another dipole; instead, it is a dipole and a net charge. If we think of the dipole as two opposite charges, the image charges are no longer of equal magnitude, and the difference is non-negligible for any choice of charges. This is a subtle issue which is detailed further in Kevin Zhou's E2 handout.

For the case of a ferromagnetic sphere, the net flux out of the sphere must be 0 by Gauss' law, so we add another charge at the center of the sphere to enforce neutrality. Then, the image of a current ring can be thought of as another current ring and a conical sheet of current from the image ring to the center of the sphere. This situation is much less tractable than intended, so we have decided to give full points for (b)-(d) to teams which realized the key idea of the sphere-plane image charge setup.

World Eater – Solution

(a) Defining $\alpha = \frac{rc^2}{GM}, \beta = \frac{\ell c}{GM}$, we have the following:

Thus, $r$ evolves under the “effective potential” $V_{\text{eff}}$. For $r \to \infty$, we have $V_{\text{eff}} \to 0$, so a faraway particle can overcome the angular momentum barrier and fall into the black hole if its “effective energy” $E = \frac{1}{2}\left(\frac{\mathrm{d}r}{\mathrm{d}\tau}\right)^2$ satisfies $E > \max(V_{\text{eff}})$.

Consider a particle with impact parameter (original distance to the axis of motion) $b$. Then, we have $\ell \approx bv$ and $E \approx v^2/2$. Let $b_m$ be the maximum impact parameter which falls into the black hole. We claim that $b_m \approx b_0$ where $b_0$ gives $\max(V_{\text{eff}}) = 0$. To see this, let $\beta_m, \beta_0$ be the corresponding values of $\beta$. Then, we can linearly approximate $\max(V_{\text{eff}}) \sim c^2(\beta_m - \beta_0)$, so $\max(V_{\text{eff}}) = E$ gives $\beta_m = \beta_0 + O(v^2/c^2) \approx \beta_0$.

Note that when $\max(V_{\text{eff}}) = 0$, the equation $V_{\text{eff}}(\alpha) = 0$ will have a double root. We have:

This has a double root when $\beta_0 = 4$, so $b_m \approx b_0 = \frac{4GM}{cv}$. Finally, the rate of mass increase is equal to the rate of mass swept out by the circle of infalling impact parameters, which is $\pi\rho v b_m^2 \approx \boxed{\frac{16\pi\rho G^2 M^2}{c^2 v}}$.

Marking Scheme: 0.25 for rewriting in terms of effective potential. 0.5 for considering particles with a given impact parameter and realizing that they will enter the black hole iff they can overcome the effective potential. 0.5 for finding the maximum impact parameter in the $v \ll c$ approximation. 0.25 for finding the rate of mass increase using the swept out circle of impact parameters.

(b) Treat the black hole's potential as Newtonian. Relativistic behavior is important for $b \sim \frac{GM}{cv}$, so we may neglect it because the Newtonian scale is $b \sim \frac{GM}{v^2}$ and $v \ll c$.

Note that all particles leave with speed $v$ in the black hole's frame, so we need to find the deflection angle $\phi(b)$. Using the properties of a hyperbola, we find $\phi = -2\arctan(a/b)$ where $a < 0$ is the semimajor axis. We have $E = -\frac{GMm}{2a}$, so $a = -\frac{GM}{v^2}$. Then, the change in the particle's axial momentum is:

Integrating over all particles:

Letting $u = 1 + (bv^2/GM)^2$:

This $\boxed{\text{diverges}}$! Clearly, the assumption that particles arbitrarily far away are affected only by the black hole is not valid.

Marking Scheme: 0.25 for stating that relativistic trajectories can be neglected. 0.75 for finding the axial impulse from a particle starting at a given impact parameter in the Newtonian approximation. Small deflections end up being the dominant contribution, so it is valid to approximate particles' trajectories as straight lines. 0.5 for integrating over all impact parameters. 0.25 for obtaining a logarithmic divergence and 0.25 for the correct prefactor.

(c) We must apply a cutoff scale $b < b_m$ to the integral. This is the distance where the gravitational influence of the black hole no longer overwhelms that of the medium; dimensionally, we have $b_m \sim (M/\rho)^{1/3}$. Then, $u_m \approx \left(b_m v^2/GM\right)^2 \sim \left(\frac{v^6}{\rho G^3 M^2}\right)^{2/3}$; because $(\rho G^3 M^2)^{1/6} \ll v$, we have $u_m \gg 1$. This gives:

Note that a multiplicative uncertainty in $b_m$ turns into an additive uncertainty in the logarithm, allowing us to obtain an asymptotic estimate.

Marking Scheme: 1.0 for realizing that a cutoff scale is needed due to assumptions about the noninteracting medium becoming invalid. Different cutoff scales (i.e. the dimensional scale presented here, or a different scale based on the Jeans' length of the black hole) may lead to different numeric prefactors; we accepted any which give the correct logarithmic term for 0.5 points.